You should know (sometimes by definition) that $(1+\frac1n)^n\to e$.
Consider the case $x\ne0$ (with $x=0$ being trivial).Then clearly $(1+\frac1n)^{nx}\to e^x$.
This already shows $(1+\frac xn)^{n}\to e^x$ at least for the subsequence of naturals $n$ where $n$ is a multiple of $x$.
For the general case, investigate the difference between $(1+\frac xn)^{n}$ and $(1+\frac 1m)^{mx}$, where $m$ is a multuple of $x$ and $m\approx \frac nx$.